Elements of Algebra

CHAP. I. OF ALGEBRA. 303 Let the number of men be $x$, and that of the women $y$, we shall then have the equation, $19x + 13y = 1000$; or $13y = 1000 - 19x = 988 + 12 - 13x - 6x$; and $y = 76 - x + \frac{12-6x}{13}$; whence it follows, that $12-6x$, or $6x - 12$, or $x - 2$, the sixth part of that number must be divisible by $13$. If, therefore, we make $x-2=13z$, we shall have $x = 13z+2$, and $y = 76 - 13z - 2 - 6z$, or $y = 74 - 19z$; which shews that $z$ must be less than $\frac{74}{19}$, and, consequently, less than $4$; so that the four following answers are possible: 1. $z=0$ gives $x=2$, and $y=74$: in which case there were 2 men and 74 women; the former paid 38 sous, and the latter 962 sous. 2. $z=1$ gives the number of men $x=15$, and that of women $y=55$; so that the former spent 285 sous, and the latter 715 sous. 3. $z=2$ gives the number of men $x=28$, and that of the women $y=36$; therefore the former spent 532 sous, and the latter 468 sous. 4. $z=3$ gives $x=41$, and $y=17$; so that the men spent 779 sous, and the women 221 sous. 9. Question 6. A farmer lays out the sum of 1770 crowns in purchasing horses and oxen; he pays 31 crowns for each horse, and 21 crowns for each ox. How many horses and oxen did he buy? Let the number of horses be $x$, and that of oxen $y$; we shall then have $31x + 21y = 1770$, or $21y = 1770 - 31x = 1764 + 6 - 21x - 10x$; or $y = 84 - x + \frac{6-10x}{21}$. Therefore $10x - 6$, and likewise its half $5x - 3$, must be divisible by $21$. If we now suppose $5x - 3 = 21z$, we shall have $5x = 21z + 3$, and hence $y = 84 - x - 2z$. But, since $x = \frac{21z+3}{5} = 4z + \frac{z+3}{5}$, we must also make $z+3=5u$; which gives $z = 5u - 3$, $x = 21u - 12$, and $y = 84 - 21u + 12 - 10u + 6 = 102 - 31u$; hence it follows, that $u$ must be greater than $0$, and yet less than $4$, which furnishes the following answers: